Question

Out of 54 randomly selected patients of a local hospital who were surveyed, 49 reported that they were satisfied with the care they recieved. Construct and interpret a 95% confidence interval for the percentage of all patients satisfied with their care at the hospital.

Answer 1

Answer:

The 95% confidence interval is given below:

$\hat{p} \pm z_{\frac{0.05}{2} } \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

$\hat{p} = \frac{49}{54} =0.9074$

$z_{\frac{0.05}{2} }=1.96$ is the critical value at 0.05 significance level

$n=54$ is the sample size

$\therefore 0.9074 \pm 1.96 \sqrt{\frac{0.9074(1-0.9074)}{54}}$

    $0.9074 \pm 0.077315056$

    $\left( 0.830, 0.985 \right)$

Therefore, the 95% confidence interval for the percentage of all patients satisfied with their care at the hospital is $\left( 0.830, 0.985 \right)$